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The amount of ROM needed to implement a 4-bit multiplier?
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Problem
For a 4-bit multiplier there are $2^4 \cdot 2^4 = 2^8$ combinations.
The output of 4-bit multiplication is 8 bits, so the amount of ROM needed is $2^8 \cdot 8 = 2048$ bits.
Why is that? Why does the ROM need all the combinations embedded into it?
What will be the case with RAM?
The output of 4-bit multiplication is 8 bits, so the amount of ROM needed is $2^8 \cdot 8 = 2048$ bits.
Why is that? Why does the ROM need all the combinations embedded into it?
What will be the case with RAM?
Solution
As shown in the figure bellow (general architecture of ROM):
to address 8-bit numbers correctly your ROM size should be: $2^8 \times 8$ bits.
to address 8-bit numbers correctly your ROM size should be: $2^8 \times 8$ bits.
Context
StackExchange Computer Science Q#9471, answer score: 8
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